Question

Eu gostaria de saber qual a integral de raiz de y^2-25/y^3 dy!

Answer 1

Calcular a integral indefinida

     $\displaystyle\int \frac{\sqrt{y^2-25}}{y^3}\,dy$


Multiplique o numerador e o denominador por y:

     $=\displaystyle\int\frac{\sqrt{y^2-25}}{y^4}\cdot y\,dy\\\\\\ =\int\frac{\sqrt{y^2-25}}{(y^2)^2}\cdot y\,dy$


Faça a seguinte substituição:

     $\sqrt{y^2-25}=u\\\\\\ \quad\Longrightarrow\quad\left\{ \begin{array}{lcl}y^2-25=u^2&\quad\Longrightarrow\quad&y^2=u^2+25\\\\ 2y\,dy=2u\,du&\quad\Longrightarrow\quad&y\,dy=u\,du \end{array} \right.$


Substituindo, a integral fica

     $\displaystyle=\int\frac{u}{(u^2+25)^2}\cdot u\,du\\\\\\ =\int \frac{u^2}{(u^2+25)^2}\,du$


Agora, faça uma substituição trigonométrica:

     $u=5\,\mathrm{tg\,}\theta\quad\ongrightarrow\quad du=5\sec^2\,d\theta$


Temos também que

     $u^2+25=(5\,\mathrm{tg\,}\theta)^2+25\\\\ u^2+25=5^2\,\mathrm{tg^2\,}\theta+25\\\\ u^2+25=25\,\mathrm{tg^2\,}\theta+25\\\\ u^2+25=25\cdot (\mathrm{tg^2\,}\theta+1)\\\\ u^2+25=25\sec^2 \theta$


Substituindo, a integral fica

     $\displaystyle=\int\frac{(5\,\mathrm{tg\,}\theta)^2}{(25\sec^2 \theta)^2}\cdot 5\sec^2\theta\,d\theta\\\\\\ =\int\frac{25\,\mathrm{tg^2\,}\theta}{625\sec^4 \theta}\cdot 5\sec^2\theta\,d\theta\\\\\\ =\int\frac{25\,\mathrm{tg^2\,}\theta}{625\sec^2 \theta\cdot \sec^2\theta}\cdot 5\sec^2\theta\,d\theta\\\\\\ =\int\frac{125\,\mathrm{tg^2\,}\theta}{625\sec^2 \theta}\,d\theta\\\\\\ =\frac{1}{5}\int\frac{\mathrm{tg^2\,}\theta}{\sec^2 \theta}\,d\theta$


Substitua tg² θ = sec² θ − 1:

     $\displaystyle=\frac{1}{5}\int\frac{\sec^2 \theta-1}{\sec^2 \theta}\,d\theta\\\\\\ =\frac{1}{5}\int\frac{\sec^2\theta}{\sec^2 \theta}\,d\theta-\frac{1}{5}\int\frac{1}{\sec^2 \theta}\,d\theta\\\\\\ =\frac{1}{5}\int 1\,d\theta-\frac{1}{5}\int\cos^2 \theta\,d\theta\\\\\\ =\frac{1}{5}\theta-\frac{1}{5}\int\cos^2 \theta\,d\theta$


Aplique uma relação para o cosseno do arco duplo:

     •  cos² θ = (1/2) · (1 + cos 2θ)


e a integral fica

     $\displaystyle=\frac{1}{5}\theta-\frac{1}{5}\int\frac{1}{2}\cdot (1+\cos 2\theta)\,d\theta\\\\\\ =\frac{1}{5}\theta-\frac{1}{10}\int 1\,d\theta-\frac{1}{10}\int\cos 2\theta\,d\theta\\\\\\ =\frac{1}{5}\theta-\frac{1}{10}\theta-\frac{1}{10}\int\cos 2\theta\,d\theta\\\\\\ =\frac{2}{10}\theta-\frac{1}{10}\theta-\frac{1}{10}\cdot \left(\frac{1}{2}\,\mathrm{sen\,}2\theta\right)+C\\\\\\ =\frac{1}{10}\theta-\frac{1}{10}\cdot \left(\frac{1}{2}\,\mathrm{sen\,}2\theta\right)+C$


Aplique a relação para o seno do arco duplo:

     •  sen 2θ = 2 sen θ cos θ


e a integral fica

     $\displaystyle=\frac{1}{10}\theta-\frac{1}{10}\cdot \left(\frac{1}{\diagup\!\!\!\! 2}\cdot \diagup\!\!\!\! 2\,\mathrm{sen\,}\theta\cos\theta\right)+C\\\\\\ =\frac{1}{10}\theta-\frac{1}{10}\,\mathrm{sen\,}\theta\cos\theta+C$


Para fazer aparecer tg θ e sec θ, multiplique e divida por cos θ:

     $\displaystyle=\frac{1}{10}\theta-\frac{1}{10}\,\frac{\mathrm{sen\,}\theta}{\cos \theta}\cdot \cos^2\theta+C\\\\\\ =\frac{1}{10}\theta-\frac{1}{10}\,\mathrm{tg\,}\theta\cdot \frac{1}{\sec^2\theta}+C$


Substitua de volta para a variável u, utilizando as relações definidas inicialmente:

     $\displaystyle=\frac{1}{10}\,\mathrm{arctg}\!\left(\frac{u}{5}\right)-\frac{1}{10}\cdot \frac{u}{5}\cdot \frac{1}{~\frac{u^2+25}{25}~}+C\\\\\\ =\frac{1}{10}\,\mathrm{arctg}\!\left(\frac{u}{5}\right)-\frac{u}{10\cdot 5}\cdot \frac{25}{u^2+25}+C\\\\\\ =\frac{1}{10}\,\mathrm{arctg}\!\left(\frac{u}{5}\right)-\frac{25}{50}\cdot \frac{u}{u^2+25}+C\\\\\\ =\frac{1}{10}\,\mathrm{arctg}\!\left(\frac{u}{5}\right)-\frac{1}{2}\cdot \frac{u}{u^2+25}+C$


Substitua de volta para a variável y:

     $\displaystyle=\frac{1}{10}\,\mathrm{arctg}\bigg(\frac{\sqrt{y^2-25}}{5}\bigg)-\frac{1}{2}\cdot \frac{\sqrt{y^2-25}}{y^2}+C\quad\longleftarrow\quad \mathsf{resposta.}$


Bons estudos! :-)